You can break the integral of a sum/difference into the sum/difference of the integrals:
[tex] \int (x^3-\cos(x))\;dx = \int x^3\; dx - \int \cos(x)\; dx [/tex]
The first integral can be solved using the power rule for integrations:
[tex] \int x^n\; dx = \frac{x^{n+1}}{n+1} \implies \int x^3\; dx = \frac{x^4}{4} [/tex]
As for the second, remember the chain of derivation of trigonometric functions:
[tex] \sin(x) \to \cos(x) \to -\sin(x) \to -\cos(x) \to \sin(x) \ldots [/tex]
The integration chain is the same, but it follows the opposite direction: the integral of the cosine function is the sine function. So, you have
[tex] - \int \cos(x)\; dx = -\sin(x) [/tex]
Since this is an indefinite integral, we must add the generic costant to the answer:
[tex] \frac{x^4}{4} -\sin(x) + C [/tex]
